📚 PDF Joiner (4) - Question 188: Integration by Substitution | PDF练习题(4)第188题：代换积分法
In this detailed walkthrough, we solve Question 188 from the Edexcel A-Level Mathematics revision collection PDF Joiner (4). The problem requires integration by substitution, one of the core skills tested in Pure Mathematics Paper 1 and Paper 2. We will break down every step, explain the underlying principles and highlight the exam techniques that can help you secure full marks on similar questions.
在这篇详尽的解题指南中，我们将解答Edexcel A-Level数学复习题集《PDF Joiner (4)》中的第188题。该题要求使用代换积分法，这是纯数试卷1和试卷2中考查的核心技能之一。我们将逐步拆解，讲解背后的原理，并强调能帮助你在类似题目中拿到满分的考试技巧。
1. Problem Statement | 题目陈述
Question 188 (Edexcel style): Use the substitution u = 2x + 1 to evaluate the indefinite integral
∫ x √2x + 1 dx
第188题（Edexcel风格）：使用代换 u = 2x + 1 计算不定积分
∫ x √2x + 1 dx
2. Understanding the Substitution Method | 理解代换积分法
Integration by substitution is essentially the reverse of the chain rule for differentiation. When we spot a composite function whose derivative appears (up to a constant multiple) in the integrand, we can set the inner function as a new variable u. This transforms a complicated integral into a simpler one in terms of u, which we then integrate before substituting back to the original variable.
代换积分法本质上是链式法则求导的逆过程。当我们发现被积函数中包含一个复合函数，且其导数（乘上一个常数）也出现在被积函数中时，就可以令内层函数为新变量 u。这样做的目的是将一个复杂的积分转化为关于 u 的简单积分，积出结果后，再代回原变量。
The general formula for indefinite integration by substitution is
∫ f(g(x)) g'(x) dx = ∫ f(u) du
In our question, the substitution is given explicitly, so we do not need to recognise the pattern — we simply follow the mechanical process.
不定积分代换法的一般公式是
∫ f(g(x)) g'(x) dx = ∫ f(u) du
在我们的题目中，代换式已直接给出，因此无需自己识别形式 — 只需遵循标准操作流程即可。
3. Choosing the Right Substitution | 选择合适的代换
The problem supplies u = 2x + 1. This is the natural choice because the expression inside the square root, 2x+1, is the 'inner function', and its derivative 2 is essentially a constant multiplier linking dx and du. Using this substitution will eliminate the square root and turn the integrand into powers of u.
本题给出 u = 2x + 1。这是一个很自然的选择，因为平方根内的表达式 2x+1 就是“内层函数”，其导数 2 在关联 dx 与 du 时仅仅是一个常数倍。使用该代换将消去根号，使被积函数完全变成 u 的幂函数。
In many examination questions, the substitution is given, but sometimes you are expected to identify it yourself. A rule of thumb is to let u be the part of the integrand that makes the integral look complicated, such as the argument of a square root, a trigonometric function or an exponential, especially when its derivative sits nearby.
在很多试题中，代换式是直接给出的，但有时也需要你自己识别。一个经验法则是：令 u 为被积函数中让积分看起来复杂的部分，比如根号内、三角函数内或指数函数的指数部分，特别是当这些部分的导数就出现在附近的时候。
4. Computing du and dx | 计算 du 与 dx
From u = 2x + 1 we differentiate with respect to x:
du/dx = 2
This can be rearranged to express dx in terms of du:
dx = du / 2
由 u = 2x + 1 对 x 求导得
du/dx = 2
整理后得到用 du 表示 dx 的式子：
dx = du / 2
We also need to write the original variable x completely in terms of u. Solving u = 2x + 1 for x yields
x = (u − 1) / 2
This step is crucial because the integrand contains an x factor outside the square root. Leaving any x in the u-integral is a very common mistake.
我们还需要将原变量 x 完全用 u 表示。由 u = 2x + 1 解出 x 得
x = (u − 1) / 2
这一步至关重要，因为被积函数中在根号外面还有一个 x 因子。许多同学的常见错误就是忘记彻底替换，在关于 u 的积分里留下了 x。
5. Rewriting the Integral | 重写积分式
Now substitute every piece into the original integral:
- Replace √2x+1 with √u.
- Replace x with (u−1)/2.
- Replace dx with du/2.
现在将所有部分代入原积分：
- 将 √2x+1 替换为 √u；
- 将 x 替换为 (u−1)/2；
- 将 dx 替换为 du/2。
The integral becomes
∫ [(u−1)/2] · √u · (du/2)
Multiplying the constants 1/2 and 1/2 gives 1/4, so we have
∫ x √2x+1 dx = ¼ ∫ (u−1) √u du
积分变为
∫ [(u−1)/2] · √u · (du/2)
常数 1/2 与 1/2 相乘得 1/4，因此得到
∫ x √2x+1 dx = ¼ ∫ (u−1) √u du
At this stage all the x-dependence has been eliminated. The integral is now entirely in the variable u.
到了这一步，所有对 x 的依赖都已被消去，积分完全处在变量 u 之下。
6. Simplifying the Integrand | 简化被积函数
Rewrite the square root as a fractional exponent: √u = u1/2. Then expand the product:
(u−1) · u1/2 = u · u1/2 − 1 · u1/2 = u3/2 − u1/2
将平方根写成分数指数的形式：√u = u1/2。然后展开乘积：
(u−1) · u1/2 = u · u1/2 − 1 · u1/2 = u3/2 − u1/2
Therefore the integral in terms of u simplifies to
¼ ∫ (u3/2 − u1/2) du
这是一个简洁的形式，能够直接积分。
7. Performing the Integration | 执行积分运算
We now integrate term by term using the power rule for integration: ∫ uⁿ du = uⁿ⁺¹/(n+1) + C, where the exponent n is not equal to −1.
现在我们使用幂函数的积分法则逐项积分：∫ uⁿ du = uⁿ⁺¹/(n+1) + C，其中指数 n 不等于 −1。
For the first term, n = 3/2, so n+1 = 5/2 and division by n+1 gives a factor of 2/5:
∫ u3/2 du = (2/5) u5/2
For the second term, n = 1/2, so n+1 = 3/2 and we get a factor of 2/3:
∫ u1/2 du = (2/3) u3/2
Putting the constant ¼ outside and including the constant of integration C yields
¼ [ (2/5) u5/2 − (2/3) u3/2 ] + C
对第一项，指数 n = 3/2，则 n+1 = 5/2，除以 n+1 得到系数 2/5：
∫ u3/2 du = (2/5) u5/2
对第二项，指数 n = 1/2，则 n+1 = 3/2，除以 n+1 得到系数 2/3：
∫ u1/2 du = (2/3) u3/2
将括号外的 ¼ 乘入并补上积分常数 C 得
¼ [ (2/5) u5/2 − (2/3) u3/2 ] + C
Multiply through by ¼:
= (1/10) u5/2 − (1/6) u3/2 + C
将 ¼ 乘进去：
= (1/10) u5/2 − (1/6) u3/2 + C
These two terms are the antiderivative in u. The expression is complete and ready for back-substitution.
这就是关于 u 的不定积分原函数，随时可以代回原变量。
8. Back-Substituting to x | 将 x 代回
Recall that u = 2x + 1. Substituting this back into our result gives the final answer in terms of x:
∫ x √2x+1 dx = (1/10)(2x+1)5/2 − (1/6)(2x+1)3/2 + C
我们记得 u = 2x + 1。将其代回结果，就得到关于 x 的最终答案：
∫ x √2x+1 dx = (1/10)(2x+1)5/2 − (1/6)(2x+1)3/2 + C
The answer can be left in this form, but it is often good practice to tidy it up. Factorise the highest common factor, which is (1/30)(2x+1)3/2:
= (1/30)(2x+1)3/2 [3(2x+1) − 5] + C
Simplify the bracket:
= (1/30)(2x+1)3/2 (6x + 3 − 5) + C = (1/30)(2x+1)3/2(6x − 2) + C
Factor a 2 from the last bracket to give a neater expression:
